J's answer to Cindy's Primary 6 Maths Speed Singapore question.
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Alternative way to do the question without any algebra or models :
From Town P to Q, the distance travelled was 120km.
From Town Q to R, the distance travelled was 135km, but in the same amount of time as from Town P to Q. That's 15km more in the same time.
And we know that the speed from Town Q to R is 10km/h faster. That means every hour, 10km more is travelled.
So if 10km more is travelled every hour, then to have travelled 15km more,
The time taken would have to be = 15km ÷ 10km x 1h = 1.5 x 1h = 1.5h
So this means that in 1.5h, 15km more was travelled from Town Q to R than from Town P to Q. And each journey took 1.5h each.
So average speed for second oart of journey = 135km ÷ 1.5h = 90km/h
From Town P to Q, the distance travelled was 120km.
From Town Q to R, the distance travelled was 135km, but in the same amount of time as from Town P to Q. That's 15km more in the same time.
And we know that the speed from Town Q to R is 10km/h faster. That means every hour, 10km more is travelled.
So if 10km more is travelled every hour, then to have travelled 15km more,
The time taken would have to be = 15km ÷ 10km x 1h = 1.5 x 1h = 1.5h
So this means that in 1.5h, 15km more was travelled from Town Q to R than from Town P to Q. And each journey took 1.5h each.
So average speed for second oart of journey = 135km ÷ 1.5h = 90km/h
Date Posted:
4 years ago
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